Planets and Satellites: Kepler's Laws — Question 2

Question 2

Equal angular sweeps do not take equal times in an eccentric orbit

For a Kepler orbit, areal velocity is constant: $dA/dt=\tfrac12r^2\dot\theta$. Its apoapsis radius is three times its periapsis radius.

1. Compare the instantaneous angular speeds at the two apsides.

2. Compare the times to sweep equal very small angles centered at those apsides.

Question 2 - Solution

Equal angular sweeps do not take equal times in an eccentric orbit

(a) Derivation and calculation

Let $h=r^2\dot\theta=2\dot A$, constant by the area law. At the apsides, $$r_p^2\dot\theta_p=r_a^2\dot\theta_a.$$ Since $r_a=3r_p$, $$\boxed{\frac{\dot\theta_p}{\dot\theta_a}=\left(\frac{r_a}{r_p}\right)^2=9}.$$

(b) Result and interpretation

For equal small angular intervals $\delta\theta$, $$\delta t=\frac{\delta A}{\dot A} \simeq\frac{r^2\delta\theta/2}{h/2}=\frac{r^2}{h}\delta\theta.$$ Thus $$\boxed{\lim_{\delta\theta\to0}\frac{\delta t_a}{\delta t_p} =\frac{r_a^2}{r_p^2}=9}.$$ For finite angular intervals, integrate $r(\theta)^2/h$ instead of treating radius as constant.

Physical check

The satellite takes longer to sweep a given small angle near apoapsis, where its areal wedge is larger.